Some Integrals Related to Catalan´s constant

In this post We will proof the following integrals related to the Catalan´s Constant. The proofs are based on the following paper: Representations of Catalan’s constant, David Bradley, 2001

$\begin{aligned} & \int_0^{1}\frac{\ln(x)}{1+x^2}\,dx=-\beta(2)\\ & \int_0^{\pi/4}\ln\left(\tan\left(x\right) \right)\,dx=-\beta(2)\\ & \int_0^{\pi/12}\ln\left(\tan\left(3x\right) \right)\,dx=-\frac13\beta(2)\\ &\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx=-\frac23\beta(2) \\ & \int_0^{2-\sqrt{3}}\frac{\ln(x)}{1+x^2}\,dx=-\frac23\beta(2) \\ & \int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx=\frac23\beta(2)-\frac{\pi}{12}\ln\left(2+\sqrt{3} \right)\\ & \int_0^{\pi/6}\frac{x}{\sin\left(x\right)}\,dx=\frac43\beta(2)-\frac{\pi}{6}\ln\left(2+\sqrt{3} \right) \end{aligned}$

$\int_0^{1}\frac{\ln(x)}{1+x^2}\,dx=-\beta(2)$ (1)

Proof:

$\begin{aligned} \int_0^{1}\frac{\ln(x)}{1+x^2}\,dx&=\int_0^{1}\ln(x)\left(\sum_{k=0}^\infty(-1)^kx^{2k} \right)\,dx\\ &=\sum_{k=0}^\infty(-1)^k\int_0^{1}x^{2k}\ln(x)\,dx\\ &=\sum_{k=0}^\infty(-1)^k\left(\frac{\ln(x)x^{2k+1}}{2k+1}\Big|_0^1-\frac{1}{2k+1}\int_0^{1}x^{2k}\,dx\right)\\ &=\sum_{k=0}^\infty(-1)^k\left(-\frac{1}{(2k+1)^2}\right)\\ &=-\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2 }\\ &=-\beta(2) \qquad \blacksquare \end{aligned}$

$\int_0^{\pi/4}\ln\left(\tan\left(x\right) \right)\,dx=-\beta(2)$ (2)

Proof:

$\begin{aligned} \int_0^{\pi/4}\ln\left(\tan\left(x\right) \right)\,dx&=\int_0^1\frac{\ln(x)}{1+x^2}\,dx &(\tan(x) \to x)\\ &=-\beta(2) \end{aligned}$

$\int_0^{\pi/12}\ln\left(\tan\left(3x\right) \right)\,dx=-\frac13\beta(2)$ (3)

Proof:

$\begin{aligned} \int_0^{\pi/12}\ln\left(\tan\left(3x\right) \right)\,dx&=\frac13\int_0^{\pi/4}\ln\left(\tan\left(x\right) \right)\,dx\\ &=-\frac13\beta(2) \qquad \blacksquare\\ \end{aligned}$

$\int_0^{\pi/12}\ln\left(\tan\left(3x\right) \right)\,dx=\frac12\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx$ (4)

Proof:

By equation (A.7) of the appendix we have

$\ln\left(\tan\left(3x\right) \right)=\ln\left(\tan\left(x\right) \right)+\ln\left(\tan\left(\frac{\pi}{3}+x\right) \right)+\ln\left(\tan\left(\frac{\pi}{3}-x\right) \right)$

Than, we can rewrite our itegral as

$\begin{aligned} \int_0^{\pi/12}\ln\left(\tan\left(3x\right) \right)\,dx&=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_0^{\pi/12}\ln\left(\tan\left(\frac{\pi}{3}+x\right) \right)\,dx+\int_0^{\pi/12}\ln\left(\tan\left(\frac{\pi}{3}-x\right) \right)\,dx\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/3}^{5\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/3}^{\pi/4}\ln\left(\tan\left(x\right) \right)\,(-dx)\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/3}^{5\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/4}^{\pi/3}\ln\left(\tan\left(x\right) \right)\,dx\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/4}^{5\pi/12}\ln\left(\tan\left(x\right) \right)\,dx\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/4}^{\pi/12}\ln\left(\tan\left(\frac{\pi}{2}-x\right) \right)\,(-dx)\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx-\int_{\pi/4}^{\pi/12}\ln\left(\cot\left(x\right) \right)\,dx\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx-\int_{\pi/4}^{\pi/12}\ln\left(\frac{1}{\tan(x)}\right)\,dx\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_{\pi/4}^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx\\ &=\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx+\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx-\int_{0}^{\pi/4}\ln\left(\tan\left(x\right) \right)\,dx\\ &=2\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx-3\int_{0}^{\pi/12}\ln\left(\tan\left(3x\right) \right)\,dx\\ &=\frac12\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx \qquad \blacksquare\\ \end{aligned}$

Equating (3) and (4) we conclude that

$\int_0^{\pi/12}\ln\left(\tan\left(x\right) \right)\,dx=-\frac23\beta(2)$ (5)

Letting $\tan(x) \to x$ in (5) we obtain

$\int_0^{2-\sqrt{3}}\frac{\ln(x)}{1+x^2}\,dx=-\frac23\beta(2)$ (6)

We can Integrate by part (6) to get

$\begin{aligned} \int_0^{2-\sqrt{3}}\frac{\ln(x)}{1+x^2}\,dx&=\ln(x)\arctan(x)\Big|_0^{2-\sqrt{3}}-\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx\\ &=\frac{\pi}{12}\ln\left(2-\sqrt{3} \right)-\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx\\ &=-\frac{\pi}{12}\ln\left(\frac{1}{2-\sqrt{3}} \right)-\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx\\ &=-\frac{\pi}{12}\ln\left(\frac{1}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}} \right)-\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx\\ &=-\frac{\pi}{12}\ln\left(2+\sqrt{3} \right)-\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx\\ \end{aligned}$

And from (6) we conclude that

$\int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx=\frac23\beta(2)-\frac{\pi}{12}\ln\left(2+\sqrt{3} \right)$ (7)

Now, let $x \to \tan\left( \frac{x}{2}\right)$ in (7)

$\begin{aligned} \int_0^{2-\sqrt{3}}\frac{\arctan(x)}{x}\,dx&=\int_0^{\pi/6}\frac{\frac{x}{2}}{\tan\left( \frac{x}{2}\right)}\sec^2\left( \frac{x}{2}\right)\,\frac{dx}{2}\\ &=\frac12\int_0^{\pi/6}\frac{x}{\frac{2\sin\left( \frac{x}{2}\right)\cos^2\left( \frac{x}{2}\right)}{\cos\left( \frac{x}{2}\right)}}\,dx\\ &=\frac12\int_0^{\pi/6}\frac{x}{\sin\left(x\right)}\,dx\\ \end{aligned}$

Ans from (7) we conclude that

$\int_0^{\pi/6}\frac{x}{\sin\left(x\right)}\,dx=\frac43\beta(2)-\frac{\pi}{6}\ln\left(2+\sqrt{3} \right)$ (8)

Appendix

$\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ (A.1)

Proof:

Recall the addition formulas of sine and cosine (see here)

$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$ (A.2)

$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ (A.3)

$\begin{aligned} \tan(x+y)&=\frac{\sin(x+y)}{\cos(x+y)}\\ &=\frac{\sin(x)\cos(y)+\sin(y)\cos(x)}{\cos(x)\cos(y)-\sin(x)\sin(y)}\\ &=\frac{\sin(x)\cos(y)+\sin(y)\cos(x)}{\cos(x)\cos(y)-\sin(x)\sin(y)}\cdot \frac{\frac{1}{\cos(x)\cos(y)}}{\frac{1}{\cos(x)\cos(y)}}\\ &=\frac{\frac{\sin(x)\cos(y)+\sin(y)\cos(x)}{\cos(x)\cos(y)}}{\frac{\cos(x)\cos(y)-\sin(x)\sin(y)}{\cos(x)\cos(y)}}\\ &=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} \qquad \blacksquare \end{aligned}$

Examples:

$\begin{aligned} \tan\left(\frac{\pi}{12}\right)&=\tan\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\\ &=\tan\left(\frac{\pi}{3}+\left(-\frac{\pi}{4}\right)\right)\\ &=\frac{\tan\left(\frac{\pi}{3}\right)+\tan\left(-\frac{\pi}{4}\right)}{1-\tan\left(\frac{\pi}{3}\right)\tan\left(-\frac{\pi}{4}\right)}\\ &=\frac{\sqrt{3}-1}{1+\sqrt{3}} \\ &=\frac{\sqrt{3}-1}{\sqrt{3}+1} \cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}\\ &=\frac{4-2\sqrt{3}}{2}\\ &=2-\sqrt{3}\qquad \blacksquare \end{aligned}$

Another two examples that will be useful in the next proof

$\begin{aligned} \tan\left(\frac{\pi}{3}-x\right) &=\tan\left(\frac{\pi}{3}+\left(-x\right)\right)\\ &=\frac{\tan\left(\frac{\pi}{3}\right)+\tan\left(-x\right)}{1-\tan\left(\frac{\pi}{3}\right)\tan\left(-x\right)}\\ &=\frac{\sqrt{3}-\tan(x)}{1+\sqrt{3}\tan(x)} \\ \end{aligned}$

$\begin{aligned} \tan\left(\frac{\pi}{3}+x\right) &=\frac{\tan\left(\frac{\pi}{3}\right)+\tan\left(x\right)}{1-\tan\left(\frac{\pi}{3}\right)\tan\left(x\right)}\\ &=\frac{\sqrt{3}+\tan(x)}{1-\sqrt{3}\tan(x)} \\ \end{aligned}$

$\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$ (A.4)

Proof:

Recall the double angle formulas (see here):

$\sin(2x)=2\sin(x)\cos(x)$ (A.5)

$\cos(2x)=\cos^2(x)-\sin^2(x)$ (A.6)

Than

$\begin{aligned} \tan(2x)&=\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}\\ &=\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)} \cdot \frac{\frac{1}{\cos^2(x)}}{\frac{1}{\cos^2(x)}}\\ &=\frac{2\tan(x)}{1-\tan^2(x)} \qquad \blacksquare \end{aligned}$

$\tan(3x)=\tan(x)\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)$ (A.7)

Proof:

$\begin{aligned} \tan\left(3x\right)&=\tan\left(2x+x\right)\\ &=\frac{\tan(2x)+\tan(x)}{1-\tan(2x)\tan(x)}\\ &=\frac{\frac{2\tan(x)}{1-\tan^2(x)}+\tan(x)}{1-\frac{2\tan(x)}{1-\tan^2(x)}\tan(x)}\\ &=\frac{2\tan(x)+\tan(x)-\tan^3(x)}{1-\tan^2(x)-2\tan^2(x)}\\ &=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}\\ &=\tan(x)\left(\frac{3-\tan^2(x)}{1-3\tan^2(x)}\right)\\ &=\tan(x)\left(\frac{\sqrt{3}+\tan(x)}{1-\sqrt{3}\tan(x)}\right)\left(\frac{\sqrt{3}-\tan(x)}{1+\sqrt{3}\tan(x)}\right)\\ &=\tan(x)\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right) \qquad \blacksquare \end{aligned}$

Reference

Representations of Catalan’s constant, David Bradley, 2001

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